You are missing the fact that
[tex]\int(\sin{x})^{3}dx[/tex]
is NOT
[tex] \frac{(\sin{x})^{4}}{4\cos{x}}[/tex]
Since there was not a cosine in the original integrand, you can't simply put one in. In particular,if you try to differentiate
[tex]\frac{(\sin{x})^{4}}{4\cos{x}} [/tex]
you do not get [itex]sin^3(x)[/itex] because you have to use the quotient rule and differentiate that new "cos(x)" also.

You surely learned how to do integrals of odd powers of sine and cosine long ago: separate out one "sin x" to use with the derivative:
[tex]\int sin^2(x) sin(x)dx= \int (1- cos^2(x)) sin(x)dx[/itex]
and let u= cos(x).

Ahhh shoot! It didn't even strike me to use substitution. I was being ignorant to the fact that it would require the quotient rule to derivate what I thought was the integral... Thank-you very much! I'm so relieved!